Configure python path using mod_wsgi

Question

I am trying to set up a simple flask app on an ec2 instance with apache2 server and mod_wsgi. Seem to be having disproportionate amount of diffculty configuring the correct python path for mod_wsgi to use.

I have placed code snippets below.

The error I get the apache2 log is:

Traceback (most recent call last):
  File "/var/www/html/flaskapp_tut/flaskapp_tut.wsgi", line 7, in <module>
    from flaskapp_tut import app as application
]  File "/var/www/html/flaskapp_tut/flaskapp_tut.py", line 1, in <module>
from flask import Flask
 ImportError: No module named flask

flask is definitely installed via anaconda installation, however clearly the wrong version of python is being used by mod_wsgi.

The log files says its using: apache/2.4.7 (Ubuntu) mod_wsgi/3.4 Python/2.7.6 configured -- resuming normal operations

However I am using python 3.x, and the anaconda installation show when I use the command "which python" ie /home/ubuntu/anaconda3/bin/python

the mod_wsgi documentation says you can configure the python path with: WSGIPythonHome /home/ubuntu/anaconda3/bin/python, however I do not know where to place this configuration.

Would appreciate any assistance. This seems as though it should be a lot more straightforard than it is, according to the steps I am using as a guide: http://www.datasciencebytes.com/bytes/2015/02/24/running-a-flask-app-on-aws-ec2/

flaskapp_tut.wsgi

#!/home/ubuntu/anaconda3/bin/python

import sys
sys.path.insert(0, '/var/www/html/flaskapp_tut')
sys.path.append('/home/ubuntu/anaconda3/bin/python')

from flaskapp_tut import app as application

flaskapp_tut.py

from flask import Flask
app = Flask(__name__)

@app.route('/')
def hello_world():
   return 'Hello from Flask!'

if __name__ == '__main__':
  app.run()

Settings within the 000-default.conf file

    DocumentRoot /var/www/html

    WSGIDaemonProcess flaskapp_tut threads=5
    WSGIScriptAlias / /var/www/html/flaskapp_tut/flaskapp_tut.wsgi

    <Directory flaskapp_tut>
        WSGIProcessGroup flaskapp_tut
        WSGIApplicationGroup %{GLOBAL}
        Order deny,allow
        Allow from all
    </Directory>

Answer 1

mod_wsgi/3.4 Python/2.7.6

That's a fairly old mod_wsgi, and what this is telling you is that the mod_wsgi you have installed is compiled against Python/2.7.6.

I recommend you get a current mod_wsgi, and make sure it's compiled against python-3.x.

Also, (and I don't think this will solve your problem, but it's worth mentioning) you can specify a python-path as an argument to WSGIDaemonProcess . This may help you get it to at least see the right stuff (and may be cleaner in some scenarios than putting that sys.path.append() in your code). See here: http://modwsgi.readthedocs.io/en/develop/configuration-directives/WSGIDaemonProcess.html .

Answer 2

It works for me in this way (Debian 9):

First , install mod-wsgi compiled for python3

sudo apt-get install libapache2-mod-wsgi-py3 

Second, add the python virtual env. lib. path in wsgi script: (start-jobs-backend.wsgi) . add also function application

 #!/usr/bin/python3
 import sys
 import os

 sys.path.insert(0, '/path/to/backend')
 sys.path.insert(0, '/path/to/venv/lib/python3.7/site-packages')

 def application(environ, start_response):
     status = '200 OK'
     output = b'Hello World! \n'
     response_headers = [('Content-type', 'text/plain'),
                ('Content-Length', str(len(output)))]


     start_response(status, response_headers)
     return [output]

 from yourproject import app as application

Third , apache conf

<VirtualHost>
    WSGIDaemonProcess backend user=www-data group=www-data threads=5 
    WSGIScriptAlias / /path/to/backend/start-jobs-backend.wsgi

    DocumentRoot  /path/to/backend
    <Files start-jobs-backend.wsgi>
        Require all granted
    </Files>
    <Directory /path/to/backend>
        AddHandler wsgi-script .wsgi
        WSGIProcessGroup backend
        WSGIApplicationGroup %{GLOBAL}
        WSGIScriptReloading On
        Options Indexes FollowSymLinks MultiViews ExecCGI
        Require all granted
    </Directory>
</VirtualHost>

I spent one week to work it out for my first flask app with wsgi deployement option, hope it can help you .

Answer 3

I recently installed mod_wsgi by building it from source

mod_wsgi Quick Configuration Guide
mod_wsgi Configuring The Source Code

I want to use Python 3.7, so I installed that (from source), and the location of my Python3.7 executable (on Ubuntu 18.04) is:

/usr/local/bin/python3.7

This location can be found by typing:

which python3.7

During the configuring of mod_wsgi, this is what I ran exactly, to make sure that my mod_wsgi is compiled against the installed Python3.7. When I initially ran just ./configure , mod_wsgi would use the Python2.7 executable by default.

./configure --with-python=/usr/local/bin/python3.7

Answer 4

If you use virtual environment highly recommend to use following format:

import logging
import sys
logging.basicConfig(stream=sys.stderr)
sys.path.insert(0, '/home/Flask/Api/')
sys.path.append('/home/Flask/fenv/lib/python3.6/site-packages/')
from run import app as application